Question

The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i$. What is $a+b$?

Answer 1

We know for our problem that the zeroes of our quadratic equation are $(2-3i)$ and $(2+3i)$, which means that the solutions for our equation are $x=2-3i$ and $x=2+3i$. We are going to use those solutions to express our quadratic equation in the form $a x^{2} +bx+c$; to do that we will use the zero factor property in reverse:
$x=2-3i$
$x-2=-3i$
$x-2+3i=0$

$x=2+3i$
$x-2=3i$
$x-2-3i=0$

Now, we can multiply the left sides of our equations:
$(x-2+3i)(x-2-3i)= x^{2} -2x-3ix-2x+4+6i+3ix-6i-3^2i^2$
= $x^{2}-4x+4-9i^{2}$
= $x^{2} -4x+4+9$
= $x^{2} -4x+13$
Now that we have our quadratic in the form $a x^{2} +bx+c$, we can infer that $a=1$ and $b=-4$; therefore, we can conclude that $a+b=1+(-4)=1-4=-3$.